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2 years ago

If R is the midpoint of QS, RS=2X-4, ST=4X-1, and RT=8X-43, find QS

Mathematics

1 answer:

Cloud [144]2 years ago

3 0

RT = RS + ST
8x - 43 = 2x - 4 + 4x - 1
8x - 43 = 6x - 5
8x - 6x = -5 + 43
2x = 38
x = 38/2
x = 19

QS = 2RS = 2(2x-4) = 2(2*19-4) = 2*34 = 68

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$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

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